# Steepest descent method - Trying to find a saddle for the integrand.

I am trying to solve the following integral using the steepest descent method:

$$I_n(\alpha)=\int_0^1x^{\alpha n} (1-x)^{\beta n} dx$$

$\beta=1-\alpha , \alpha > 0, \beta > 0$.

This is from a book, and the solution says that the saddle point is at $x=\alpha$. I cannot see why. I tried to put it in the form: $$\int_a^be^\frac{-f(x)}{\lambda}dx$$

To do this, I used $x^{\alpha n}=e^{\alpha n \ln(x)}$ and $(1-x)^{\beta n} = e^{\beta n\ln(1-x)}$. After simplifying, this gave me $\lambda=\frac{1}{n}$ and $f(x)=\ln(x^\alpha(1-x)^{1-\alpha})$. To minimise $f(x)$, I don't see how $x$ can equal $\alpha$. Unfortunately, the book doesn't give more information on how the saddle point is at $x=\alpha$. I would like to know how to obtain this result and complete the integral.

• Comment to the post (v1): Presumably $n$ is assumed large. Nov 17 '17 at 15:22

For any $n\in\mathbb{N}$ and $\alpha\in(0,1)$ we have:
$$\begin{eqnarray*} I_n(\alpha)=\int_{0}^{1}\left[x^\alpha(1-x)^{1-\alpha}\right]^n\,dx &=& B(n\alpha+1,n-n\alpha+1)\\&=&\frac{\Gamma(n\alpha+1)\,\Gamma(n-n\alpha+1)}{\Gamma(n+1)}\end{eqnarray*}$$ by Euler's Beta function. Asymptotics can be derived from Gautschi's inequality and/or Stirling's approximation. As an alternative, the maximum value of $f(x)=x^\alpha(1-x)^{1-\alpha}$ over $[0,1]$ is attained at $x=\alpha$, and $f''(\alpha)=-\frac{1}{\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha$. It follows that
$$I_n(\alpha)\approx \alpha^{n\alpha}(1-\alpha)^{n-n\alpha}\int_{-\infty}^{+\infty}e^{-\frac{n}{2\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha z^2}\,dz=\alpha^{n\alpha}(1-\alpha)^{n-n\alpha}\sqrt{\frac{2\pi}{\frac{n}{\alpha}\left(\frac{\alpha}{1-\alpha}\right)^\alpha}}.$$